﻿/*
Easy as A+B 
Time Limit:1000MS  Memory Limit:32768K


Description:
These days, I am thinking about a question, how can I get a problem as easy as A+B? 
It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.
 Give you some integers, your task is to sort these number ascending.
 You should know how easy the problem is now! Good luck!

Input:
Input contains multiple test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains an integer N (1<=N<=1000 the number of integers to be sorted) and then N integers follow in the same line. It is guarantied that all integers are in the range of 32bit-int. 
Output:
For each case, print the sorting result, and one line one case. 
Sample Input:
2
3 2 1 3
9 1 4 7 2 5 8 3 6 9
Sample Output:
1 2 3
1 2 3 4 5 6 7 8 9
*/
#include <algorithm>
#include <iostream>
#include <vector>
#include <iterator>
using namespace std;

int main()
{
	int n;
	cin>>n;
	while (n--)
	{
		int t;
		cin>>t;
		vector<int> v;
		v.reserve(t);

		for (int num;t--;v.push_back(num))
		{			
			cin>>num;			
		}
		sort(v.begin(), v.end());
		copy(v.begin(), v.end()-1, ostream_iterator<int>(cout, " "));
		cout<<v.back()<<endl;
	}
	return 0;
}